91Ó°ÊÓ

(a) The quantity of electron beam that travels through a closed surface is known as the electric flux. The electric field through a surface is related to the charge inside the layer, according to Gauss's law. Because the electric field is homogeneous here, we can compute the electric flux ${\mathrm{Φ}}_E\text{by}$using equation $(24.3)$

${\mathrm{Φ}}_e=EA$

Because the blanket has two different sides, we must double the flow by twice to get the transmit.

${\mathrm{Φ}}_e=2EA$

In addition, the electric field is proportional to the charge $Q_{in}$

${\mathrm{Φ}}_e=\frac{Q_{\text{in}}}{{\mathrm{Î\mu }}_o}$

Because the left edges of equations $\left(1\right)and\left(2\right)$are one and the same, we can state every next calculation in almost the same way.

$\begin{array}{r}2EA=\frac{Q_{\text{in}}}{{\mathrm{Î\mu }}_o}\\ 2EA=\frac{\mathrm{ÒÏ}V}{{\mathrm{Î\mu }}_o}\end{array}$

The thickness of the sheet is " because it is placed among $-Z$and $Z$

$h=z−(−z)=2z$

We consider the area of the sheet is $A$, so the volume of the sheet is

$V=hA=2zA$

.

To get $E$, plug the formula of $V$into equation $\left(3\right)$.

$2EA=\frac{\mathrm{ÒÏ}V}{{\mathrm{Î\mu }}_o}$

$2EA=\frac{\mathrm{ÒÏ}\left(2zA\right)}{{\mathrm{Î\mu }}_o}$

$E=\frac{\mathrm{ÒÏ}z}{{\mathrm{Î\mu }}_o}$

(b) Since this gaussian surface well outside plate seems to be between and , the module's breadth is '

$h_o=z_o−\left(−z_o\right)=2z_o$

We take the frame's area to be $A$, hence the sheet's volume is

$V=h_{o}A=2z_{o}A$

Plug the expression of $V$into equation (3) to get $E$by

$2EA=\frac{\mathrm{ÒÏ}V}{{\mathrm{Î\mu }}_o}$

$2EA=\frac{\mathrm{ÒÏ}\left(2_{o}zA\right)}{{\mathrm{Î\mu }}_o}$

$E=\frac{\mathrm{ÒÏ}{z}_o}{{\mathrm{Î\mu }}_o}$

(c) The electric fie approaches. The electric field remains constant as the distance $z$rises, as shown in parts (a) and (b). To draw the figure, we draw a straight line with a positive slope till point $z_0$, and the electric field is constant after $z_0$, as shown in the diagram below.